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valina [46]
2 years ago
10

Please quickly! Out of 5,500 people who attended the football game, fifty-six percent were rooting for the home team. How many p

eople rooted for the home team?
Mathematics
1 answer:
tigry1 [53]2 years ago
5 0
3,080 people rooted for the hometeam.
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3 years ago
A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ
jonny [76]

Let \vec r(t),\vec v(t),\vec a(t) denote the rocket's position, velocity, and acceleration vectors at time t.

We're given its initial position

\vec r(0)=\langle0,0,10\rangle\,\mathrm m

and velocity

\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}

Immediately after launch, the rocket is subject to gravity, so its acceleration is

\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}

where g=9.8\frac{\rm m}{\mathrm s^2}.

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}

\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}

(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}

and

\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m

\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m

\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}

b. The rocket stays in the air for as long as it takes until z=0, where z is the z-component of the position vector.

10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}

c. The rocket reaches its maximum height when its vertical velocity (the z-component) is 0, at which point we have

-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)

\implies\boxed{z_{\rm max}=125,010\,\rm m}

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Use the equation
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1/5+ (-4) is the same as 1/5 - 4
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